Question

QUESTION 1:

An SRS of 450 high school seniors gained an average of x¯¯¯x¯ = 21 points in their second attempt at the SAT Mathematics exam. Assume that the change in score has a Normal distribution with standard deviation 53.

Find a 90% confidence interval for μμ based on this sample.

Confidence interval (±±0.01) is between  and

What is the margin of error (±±0.01) for 90%?

Suppose we had an SRS of just 100 high school seniors. What would be the margin of error (±±0.01) for 90% confidence?

QUESTION 2:

The National Assessment of Educational Progress ( NAEP) includes a mathematics test for eigth-grade students. Scores on the test range from 0 to 500. Suppose that you give the NAEP test to an SRS of 900 8th-graders from a large population in which the scores have mean 262 and standard deviation 117. The mean x¯¯¯x¯ will vary if you take repeated samples.

The sampling distribution of x¯¯¯x¯ is approximately Normal. It has mean 262.

What is its standard deviation (±±0.001)?

Answer 1

Answer 1

(A) using TI 84 calculator

pres stat then tests then Zinterval

enter the data

sigma = 53

xbar = 21

n = 450

c-level = 0.90

press calculate,we get

(16.89, 25.11)

Margin of error = (upper limit - lower limit)/2

= (25.11 -16.89)/2

= 4.11

Using n = 100 with alpha = 1-0.90 = 0.10

Margin of error = CONFIDENCE(alpha,sd,size)

=CONFIDENCE(0.10,53,100)

= 8.72

Answer 2

Standard deviation = sd/sqrt(n)

where sd = 117

and n is sample size = 900

so, we get

standard deviatio = 117/sqrt(900)

= 3.900